Sum and product puzzle: Difference between revisions

using filters would be much slower in Julia, which often favors fast for loops over lists for speed.

<lang Julia>

primeslt100list = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,

47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

for i in 2:Int(floor(x/2))

return false

end

true

end

 

function satisfies2(x)